1

给定以下 MySQL 表结构：

CREATE TABLE `order_params`( `order_id` BIGINT(30) NOT NULL, 
              `key` VARCHAR(50) NOT NULL, `value` VARCHAR(255) NOT NULL );

而这个数据：

INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('1', 'browser', 'Firefox'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('1', 'os', 'Windows'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('2', 'browser', 'Firefox'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('2', 'os', 'Windows'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('3', 'browser', 'Firefox'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('3', 'os', 'OSX'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('4', 'browser', 'Safari'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('4', 'os', 'OSX'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('5', 'browser', 'Safari'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('5', 'os', 'OSX'); 
INSERT INTO `order_params` (`order_id`, `key`, `value`) VALUES ('5', 'version', '5');

如何获得以下结果？

browser Firefox os Windows              2
browser Firefox os OSX                  1
browser Safari os OSX                   1
browser Safari os OSX version 5         1

右边的数字是匹配唯一键/值组合的记录数。这甚至可能吗？

好的，更新以表明我已经尝试过：

SELECT CONCAT(`key`, `value`), COUNT(*)
FROM order_params 
GROUP BY `order_id`, `key`, `value`;

这是结果：

browserFirefox  1
osWindows       1
browserFirefox  1
osWindows       1
browserFirefox  1
osOS X          1
browserSafari   1
osOS X          1

我也试过这个：

SELECT `key`, `value`, COUNT(*)
FROM order_params 
GROUP BY `key`, `value`;

产生这个：

browser Firefox  3
browser Safari   1
os OS X          2
os Windows       2

显然，这些都不是我们想要的结果。

4

1 回答 1

3

一种方法是两个聚合阶段

select browser, os, version, count(*)
from (select order_id,
             max(case when `key` = 'browser' then `value` end) as browser,
             max(case when `key` = 'os' then `value` end) as os,
             max(case when `key` = 'version' then `value` end) as version
      from order_params op
      group by order_id
     ) p
group by browser, os, version

如果你真的想要你拥有的字符串，你可以将它们连接在一起：

select concat(coalesce(concat('browser ', browser), ''),
              coalesce(concat('os ', os), ''),
              coalesce(concat('version ', version), ''), count(*)
from (select order_id,
             max(case when `key` = 'browser' then `value` end) as browser,
             max(case when `key` = 'os' then `value` end) as os,
             max(case when `key` = 'version' then `value` end) as version
      from order_params op
      group by order_id
     ) p
group by os, browser, version
于 2013-03-29T21:28:11.320 回答